Asked by Precious
A wheel rotating with an angular velocity of 16rad/sec is slowed to an angular velocity of 4rad/sec by an angular acceleration of 2rad/sec^2.find
a. The angle turned in the process of slowing down
b. The time required
a. The angle turned in the process of slowing down
b. The time required
Answers
Answered by
Henry
a. V^2 = Vo^2 + 2a*d.
d = (V^2-Vo^2)/2a.
d = (16-256)/-4 = 60 Rad.
b. t = (V-Vo)/a = (4-16)/-2 = 6 s.
d = (V^2-Vo^2)/2a.
d = (16-256)/-4 = 60 Rad.
b. t = (V-Vo)/a = (4-16)/-2 = 6 s.
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